$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $

If the system of linear equation $x + 2ay + az = 0,$ $x + 3by + bz = 0,$ $x + 4cy + cz = 0$  has a non zero solution, then $a,b,c$

$\left| {\,\begin{array}{*{20}{c}}{{b^2} - ab}&{b - c}&{bc - ac}\\{ab - {a^2}}&{a - b}&{{b^2} - ab}\\{bc - ac}&{c - a}&{ab - {a^2}}\end{array}\,} \right| = $

The determinant $\left| {\begin{array}{*{20}{c}}{1\, + \,a\, + \,x}&{a\, + \,y}&{a\, + \,z}\\{b\, + \,x}&{1\, + \,b\, + \,y}&{b\, + \,z}\\{c\, + \,x}&{c\, + \,y}&{1\, + \,c\, + \,z}\end{array}} \right|$ $=$

Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If

$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$

then value of $\lambda^{2}$ is equal to $.....$

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$